Largest ellipse with fixed eccentricity inscribed in a hyperbola

Calculus Level pending

Consider the hyperbola $y^2 - x^2 = 1$ , you want to inscribe the largest possible ellipse with major to minor axes ratio of $2$ , such that the ellipse touches the hyperbola at its vertex $(0,1)$ but is otherwise above $y=1$ . The area of the largest possible ellipse with these specifications is $x \pi$ , where $x$ is a positive integer. Find $x$ .

The answer is 8.

1 solution

Nick Kent
May 8, 2021

The equation of upper half of the hyperbola:

$y = f(x) = \sqrt{(1+{x}^{2})}$

Then the curvature at point $(0, 1)$ is:

$\kappa(x=0) = \frac{f''}{{(\sqrt{1+f'})}^{3}} = \frac{1}{{(\sqrt{1+0})}^{3}} = 1$

To get the largest ellipse, it should touch the hyperbola at the "curviest" point - the vertex. The curvature of the vertex is $\frac{{b}^{2}}{a}$ . So:

$\frac{{b}^{2}}{a} = 1$

Since $\frac{a}{b} = 2$ , $a = 4$ and $b = 2$ . The resulting maximum area is $\boxed{8\pi}$ .

Largest ellipse with fixed eccentricity inscribed in a hyperbola

The answer is 8.

1 solution

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